Calculus III; standard form of & angle to eliminate the cross produc

In summary: I am sorry if you didn't understand me before. So the correct result would be u^2/4+v^2/12=1 ?In summary, the conversation discusses the process of solving a problem involving a change of variables and finding the standard form of an ellipse in terms of the new variables. The final result is u^2/4 + v^2/12 = 1.
  • #1
Molderish
11
0
Hi; I've been trying to solve the problem myself but i really don't what could be wrong;

The problem says :

Make the change of variables
x=ucos−vsin
y=usin+vcos

where the angle 0<(phi)<2 is chosen in order to eliminate the cross product term in
x^2+xy+y^2=6

Then find the standard form of equation in the (uv) variables. (Enter a function of (uv).)

---------------=1

well what I've found the angle is (pi/4) which would eliminates the cross product terms "uv" when i make the substituion..

then I've tried to reorder the equation: x^2+xy+y^2=6 which is an elipse with center at (0,0) ; semimajor axis=2sqrt3 & semiminor axis=2

then i got the equation (x^2/12)+(y^2/4)=1 then change variables again and i got ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4.
its incorrect.

If you could make a step by step solution , would be great , thanks in advanced.
 
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  • #2
Molderish said:
Hi; I've been trying to solve the problem myself but i really don't what could be wrong;

The problem says :

Make the change of variables
x=ucos−vsin
y=usin+vcos

where the angle 0<(phi)<2 is chosen in order to eliminate the cross product term in
x^2+xy+y^2=6

Then find the standard form of equation in the (uv) variables. (Enter a function of (uv).)

---------------=1

well what I've found the angle is (pi/4) which would eliminates the cross product terms "uv" when i make the substituion..

then I've tried to reorder the equation: x^2+xy+y^2=6 which is an elipse with center at (0,0) ; semimajor axis=2sqrt3 & semiminor axis=2

then i got the equation (x^2/12)+(y^2/4)=1 then change variables again and i got ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4.
its incorrect.

If you could make a step by step solution , would be great , thanks in advanced.
We don't provide step by step solutions at PF.

Why does the equation you get,

(x2/12)+(y2/4)=1,

have x & y in it, rather than u & v ?



To do the integration with u,v, don't forget your Jacobian -- although in this case it's a convenient value.
 
  • #3
Because what i tried to do its to re-write the equation " x^2+xy+y^2=6" which i found easier. then i plugged into it the "x" and "y" the problems gives.

but what i am not really sure if it means "elipse standard form"; i think the angle i got it's right since " ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4." & "x^2+xy+y^2=6" are the same elipse.. but i don't why its incorrect.
 
  • #4
Molderish said:
Because what i tried to do its to re-write the equation " x^2+xy+y^2=6" which i found easier. then i plugged into it the "x" and "y" the problems gives.

but what i am not really sure if it means "elipse standard form"; i think the angle i got it's right since " ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4." & "x^2+xy+y^2=6" are the same elipse.. but i don't why its incorrect.
Sure, you have the correct angle.

plugging in u & v, I get

u2/4 + v2/12 = 1 ,

which is the ellipse you describe, with major axis along the v axis.
 
  • #5
but wouldn't it be x=ucos−vsin--> so ((ucos(pi/4)-vsin(pi/4))^2)/12 & y=usin+vcos----> ((usin(pi/4))+(vcos(pi/4)))^2/4?? did you reduce factors?
 
  • #6
The presence of the xy term is what causes the ellipse to be rotated with respect to the x and y axes. You're trying to find a rotated set of coordinates in which the ellipse isn't rotated. You don't want the ellipse to look exactly the same in both sets of coordinates.
 
  • #7
oh ok! i see.. well i think is a minor detail i couldn't see , thanks so much!
 
  • #8
Molderish said:
but wouldn't it be x=ucos−vsin--> so ((ucos(pi/4)-vsin(pi/4))^2)/12 & y=usin+vcos----> ((usin(pi/4))+(vcos(pi/4)))^2/4?? did you reduce factors?

Take the original equation

x2+xy+y2 = 6

Substitute as directed to express it in terms of u & v.

As a result of that, I got u2/4 + v2/12 = 1


It looks like you may have done that, then switched back to x & y , because using u & v was unfamiliar to you. -- then you tried the substitution again into that result.

Is that what you did ?
 
  • #9
yeah actually i make the change of variables & i end up.. with "uv" terms.

then i found the angle to make them 0. because it was the cross product terms then i just got a u^2 and v^2 with sin & cos =6 at this point i didn't know how to reorder the equation.

that's why i go back to the original equation. i though you needed the "angle" to make the direct substituion and have the values of cosines and sines once you have the "standard form of the given elipse".
 
  • #10
Molderish said:
yeah actually i make the change of variables & i end up.. with "uv" terms.

then i found the angle to make them 0. because it was the cross product terms then i just got a u^2 and v^2 with sin & cos =6 at this point i didn't know how to reorder the equation.
Your result at this point is what is asked for in the original problem: a function of (u,v) .

that's why i go back to the original equation. i though you needed the "angle" to make the direct substituion and have the values of cosines and sines once you have the "standard form of the given elipse".
 
  • #11
SammyS said:
Your result at this point is what is asked for in the original problem: a function of (u,v) .

Actually at first a tried with this result. but it said ,it was "incorrect" i guess because i didn't reorder the equation.
 

FAQ: Calculus III; standard form of & angle to eliminate the cross produc

1. What is the standard form of a vector in Calculus III?

The standard form of a vector in Calculus III is a vector in the form of ai + bj + ck, where a, b, and c are the components of the vector in the x, y, and z directions, respectively.

2. How is the standard form of a vector used to eliminate the cross product?

The standard form of a vector is used to eliminate the cross product by identifying the i, j, and k components of each vector and using them to find the determinant of a 3x3 matrix. The resulting value is the scalar that eliminates the cross product and allows for the computation of the angle between the vectors.

3. What is the angle between two vectors in Calculus III?

The angle between two vectors in Calculus III is the angle formed by the two vectors at their intersection point. This angle can be calculated using the dot product or the cross product of the two vectors.

4. How is the angle between two vectors related to the standard form and cross product?

The angle between two vectors is related to the standard form and cross product through the use of the dot product or cross product formulas. The dot product formula uses the magnitudes and cosine of the angle to compute the dot product, while the cross product formula uses the magnitudes and sine of the angle to compute the cross product.

5. What is the significance of eliminating the cross product in Calculus III?

Eliminating the cross product in Calculus III allows for the computation of the angle between two vectors, which is an important concept in vector analysis. It also simplifies the computation of other vector operations, such as finding the projection of one vector onto another or finding the area of a parallelogram formed by two vectors.

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